摘要： int AbstractFile::OpenFile(const char* pathName, int oflag, mode_t mode){ m_fileID = open(pathName, oflag, mode); if (m_fileID == -1) { m_fileID = 0; // error message perror("###########OPEN FILE Error,msg:"); return ERROR_OPEN_FILE; } return m_fileID;}---一下就定位出来出错的原因： too many open file然后根据此文章http: 阅读全文